The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:

(1) $\frac {1}{2}$ (2) $\frac {1}{2\sqrt 2}$

(3) $\frac {2}{3}$ (4) $\frac {2}{3\sqrt 2}$

(3) $\frac {2}{3}$ (4) $\frac {2}{3\sqrt 2}$

*Solution*Activity $A \propto N = \frac {N_0}{2^n}$

$A_0 \propto N_0 $

Fraction of original activity $\frac {A}{A_0}=\frac {1}{2^n}$, where n = number of half-lives

If half-life is 100 hours, number of half-lives in 150 hours = 1.5 = n

So, $\frac {A}{A_0}=\frac {1}{2^{1.5}}=\frac {1}{2\sqrt 2}$

Answer: (2)